p0504

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Unformatted text preview: 5.4. CHAPTER 5, PROBLEM 4 303 5.4 Chapter 5, Problem 4 Problem: A block of mass m1 = m is moving to the right with speed v1 = V and a block of mass m2 = 2m is moving to the left with speed v2 = 1 V . The coefficient of restitution is e. Ignoring effects 2 of friction, determine the velocity vectors for the blocks immediately after impact, v1 and v2 , in terms of V and e. Compute the velocity vectors for e = 0.4 and V = 6 m/sec. ............................................... ... . ........ ......... . . ..... ..................................................... . . . . . . . . . ........ .. .......................................................... . . .......................................................... . .. . .......................................................... . . . .......................................................... . .......................................................... . . . . .......................................................... . . .......................................................... . .......................................................... .......................................................... . . . . .......................................................... . . .......................................................... .......................................................... . . ........................... . ...... . . . . ....................... ... . . . . . .. .......................................................... . .......................................................... .............................. ............................. .................................2.................... . . . . .. . . . . . .. . . . . . ............................. ................................. ........2.................................................................. .......................................................... ............................. ............... ............ . ...........................................1............ . ............................................................................. . . .. ... . . .. .. . .. ..... ..... . . . . . . ............................. ................................ . 1. .. . . .......................................................... ............................. ............................. 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... ... ... ... ... ... ... ... .. . . . .. ... ... ... ... ... .. . . . . . . . . .. ... ... .. .. .. . . . . . . . . . . . .. ... .. . . . . . . . . . . . .. . .......................... . . . ................... . . . . . . . . .......... .... . . . . . . . . . . . ....... . . . . . . . . . . . . . .. ................................................................................................................................................................. .... ......... . ............ ............. m v v m Solution: This is a direct central impact with the line of impact for the blocks being the x axis. Momentum Conservation. Using the fact that the initial velocity components for Blocks 1 and 2 are v1 = V and v2 = − 1 V , while m1 = m and m2 = 2m, momentum conservation along the line of impact 2 tells us that 1 mV − 2m V = mv1 + 2mv2 2 Dividing through by m yields v1 + 2v2 = 0 Impact Relation. For a coefficient of restitution equal to e, we have v2 − v1 = e (v1 − v2 ) = 3 eV 2 To complete the solution, we add the equations resulting from momentum conservation and the impact relation, which yields 3 1 =⇒ v2 = eV 3v2 = eV 2 2 Substituting for v2 in the momentum-conservation equation tells us that v1 = −2v2 Therefore, the velocity vectors after the impact are v1 v2 = −eV i 1 = eV i 2 =⇒ v1 = −eV For the given values of V = 6 m/sec and e = 0.4, we have v1 = −0.4(6 m/sec) = −2.4 m/sec v2 = 0.5(0.4)(6 m/sec) = 1.2 m/sec Therefore, the velocity vectors are v1 = −2.4 i and v2 = 1.2 i (m/sec) ...
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